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===== Supermartingale convergence results =====
 --- //[[rdouc909@gmail.com|randal]] 2021/03/30 14:04//

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**__Theorem__**. 
Let $\mcf=(\mcf_n)_{n\in\nset}$ be a filtration and let $\seq{X_n}{n\in\nset}$ be an $\mcf$-adapted sequence of $\lone$-random variables such that 
  *  $M=\sup_{n\in\nset} \PE[(X_n)^-]<\infty$
  *  for all $n\geq 1$, we have $\PE[X_{n}|\mcf_{n-1}]\leq X_{n-1}$.   
that is, $\seq{X_n}{n\in\nset}$ is a $(\mcf_n)_{n\in\nset}$-supermartingale, with negative part bounded in $\lone$. 
Then, almost surely, $X_\infty=\lim_{n\to\infty} X_n$ exists and is in $\lone$.

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==== Proof ====

Let $a<b$ and define  $C_1=\indiacc{X_0<a}$ and for $n\geq 2$, 
$$
C_n=\indiacc{C_{n-1}=1,X_{n-1} \leq b}+\indiacc{C_{n-1}=0,X_{n-1} < a}
$$
In words, the first time $C_n $ flags $1$ is when $X_{n-1}<a$. Then it flags $1$ until $X_{n-1}$ goes above $b$. Then it flags $0$ until $X_n$ goes below $a$. So consecutive sequences of $C_n=1$ are linked with upcrossings of $[a,b] $ for $(X_n) $. Now, define 
$$
Y_n=\sum_{k=1}^n C_k(X_k-X_{k-1}).
$$

Define $U_N[a,b]$ the number of upcrossings of $[a,b]$ for $(X_n)_{0\leq n \leq N } $. Then, 
$$
Y_N=\sum_{k=1}^N C_k(X_k-X_{k-1}) \geq (b-a) U_N[a,b]-(X_N-a)^-
$$
From the fact that $\seq{X_n}{n\in\nset}$ is a $\mcf$-supermartingale and $(C_n)_{n\in\nset}$ is $\mcf$-previsible, we deduce that $\seq{Y_n}{n\in\nset}$ is also a $\mcf$-supermartingale so that $\PE[Y_N]\leq 0$. Then,  
$$
(b-a) \PE[U_N[a,b]] \leq \PE[(X_N-a)^-]=\PE[(a-X_N)^+] \leq a^+ + \PE[(-X_N)^+]\leq a^+ + M
$$
Letting $N$ goes to infinity, the monotone convergence theorem yields: 
$$
(b-a) \PE[U_\infty[a,b]] \leq  a^+ + M
$$
and thus, $\PP(U_\infty[a,b]<\infty)=1$ for all $a<b$. Now, 
\begin{align*}
  \PP(\liminf_n X_n<\limsup_n X_n) &\leq \sum_{a,b\in \mathbb{Q}, a<b} \PP\lr{\liminf_n X_n <a<b<\limsup_n X_n}\\
  & \leq \sum_{a,b\in \mathbb{Q}, a<b}  \PP(U_\infty[a,b]=\infty)=0
\end{align*}
which shows that $X_\infty=\lim_{n \to \infty} X_n$ exits almost surely.

Moreover, $\PE[X_0] \geq \PE[X_n]=\PE[X^+_n]-\PE[X^-_n]$ so that
$$
\sup_n \PE[|X_n|]=\sup_n \lr{\PE[X^+_n]+\PE[X^-_n]}  \leq \PE[X_0] +2 \sup_n \PE[X^-_n] \leq \PE[X_0] +2 M<\infty  
$$
which implies by Fatou's lemma that $\PE[|X_\infty|]=\PE[\liminf_{n}|X_n|] \leq \liminf_{n} \PE[|X_n|] \leq \sup_n   \PE[|X_n|]<\infty$. The proof is completed.
===== Corollary: Submartingale convergence results =====
 
{{anchor:submartingale:}}
As a consequence, the conclusion also holds if $\seq{X_n}{n\in\nset}$ is a $\mcf$-submartingale, with positive part bounded in $\lone$ : indeed, we only need to apply the previous result to $-X_n$.

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**__Corollary__** Assume that $\mcf=(\mcf_n)_{n\in\nset}$ is a filtration and let $\seq{X_n}{n\in\nset}$ be an $\mcf$-adapted sequence of $\lone$ random variables such that 
  * $M=\sup_{n\in\nset} \PE[(X_n)^+]<\infty$
  * for all $n\geq 1$, we have $\PE[X_{n}|\mcf_{n-1}]\geq X_{n-1}$.
that is, $\seq{X_n}{n\in\nset}$ is a $\mcf$-submartingale, with positive part bounded in $\lone$. 
Then, almost surely, $X_\infty=\lim_{n\to\infty} X_n$ exists and is in $\lone$.

</WRAP>

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